Question 16.8.1: For all values of p, find the solutions of the system px + y......
For all values of p, find the solutions of the system
\begin{matrix} px + y \ \ \ \quad \quad = 1 \\ \ \ \ \ \ x − y + z = 0 \\ \ \quad \quad 2y − z = 3 \end{matrix}The coefficient matrix has determinant
\left|A\right| =\left | \begin{matrix} p & \quad 1 & \quad 0 \\ 1 & -1 & \quad 1 \\ 0 & \quad 2 & -1 \end{matrix} \right | =1-pAccording to Theorem 16.8.1, the system has a unique solution if 1-p\neq0 —that is, if p\neq1. In this case, the determinants in (16.8.2) are
D_{j}=\left | \begin{matrix} a_{11} & … & a_{1,j-1} & b_{1} & a_{1,j+1} & … & a_{1n} \\ a_{21} & … & a_{2,j-1} & b_{2} & a_{2+j+1} & … & a_{2n} \\ . & & . & . & . & & . \\ : & & : & : & : & & : \\ a_{n1}&…& a_{n, j−1}& b_{n}& a_{n, j+1} & …& a_{nn}\end{matrix} \right | (16.8.2)
D_{1} =\left | \begin{matrix} 1 & 1 & 0 \\ 0 & -1 & 1 \\ 3 & 2 & -1 \end{matrix} \right | , D_{2} =\left | \begin{matrix} p & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 3 & -1 \end{matrix} \right | \quad \quad \text{and}\quad \quad D_{3} =\left | \begin{matrix} p & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 2 & 3 \end{matrix} \right |
whose numerical values are D_{1} = 2, D_{2} = 1 − 3p, and D_{3} = −1 − 3p. Then, for p \neq 1, Eq. (16.8.4) yields
x_{1}={\frac{D_{1}}{|\mathbf{A}|}}\,,\,\,\,x_{2}={\frac{D_{2}}{|\mathbf{A}|}}\,,\cdot\cdot\cdot\,,\,\,\,x_{n}={\frac{D_{n}}{|\mathbf{A}|}} (16.8.4)
x=\frac{D_{1}}{|{\bf A}|}=\frac{2}{1-p},\quad y=\frac{D_{2}}{|{\bf A}|}=\frac{1-3p}{1-p}, \mathrm{and}\;\;\;\;\;z=\frac{D_{3}}{|{\bf A}|}=\frac{-1-3p}{1-p}On the other hand, in case p = 1, the first equation becomes x + y = 1. Yet adding the last two of the original equations implies that x + y = 3. There is no solution to these two contradictory equations in case p = 1.^{9}
^{9} It might be instructive to solve this problem by using Gaussian elimination, starting by interchanging the first two equations.